A326825 a(n) is the number of iterations needed to reach 1 or 5 starting at n and using the map k -> (k/2 if k is even, otherwise k + (smallest triangular number > k)). Set a(n) = -1 if the trajectory never reaches 1 or 5.

0, 1, 6, 2, 0, 7, 7, 3, 5, 1, 12, 8, 10, 8, 8, 4, 6, 6, 4, 2, 15, 13, 58, 9, 56, 11, 52, 9, 36, 9, 15, 5, 28, 7, 13, 7, 20, 5, 18, 3, 18, 16, 16, 14, 59, 59, 59, 10, 14, 57, 57, 12, 55, 53, 51, 10, 22, 37, 55, 10, 24, 16, 22, 6, 35, 29, 14, 8, 27, 14, 12, 8, 10, 21, 23, 6, 27, 19

Offset: 1

Author

Ali Sada, Oct 20 2019

Keywords

• nonn

Comments

It is conjectured that this algorithm will always terminate at 1 or 5.

Jim Nastos verified the conjecture for n <= 354999.

The conjecture holds for all n <= 10^9. - Jon E. Schoenfield, Oct 20 2019

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1000, May 08 2025

Crossrefs

Cf. A006577, A326823 (similar with squares instead of triangular numbers).

Programs

PARI

M326825=Map([1, 0; 5, 0]); apply( {A326825(n)=if(mapisdefined(M326825, n, &n), n, mapput(M326825, n, 1+n=A326825(if(n%2, n+binomial((sqrtint(8*n+8)+3)\2, 2), n\2))); 1+n)}, [1..77]) \\ M. F. Hasler, May 08 2025

Formula

For n = 11: 11+15 = 26; 26/2 = 13; 13+15 = 28; 28/2 = 14; 14/2 = 7; 7+10 = 17; 17+21 = 38; 38/2 = 19; 19+21 = 40; 40/2 = 20; 20/2 = 10; 10/2 = 5; taking 12 steps to reach 5, so a(11) = 12.

Extensions

Edited by Jon E. Schoenfield and N. J. A. Sloane, Oct 20 2019